Determining Weights of Terms in ODEs of General Order n

[Benny]

I was reading something about ODEs and I came across a section which discusses the generalisation of first-order isobaric equations to equations of general order n. The definition I have is that an n-th order isobaric equation is one in which every term can be made dimensionally consistent upon giving y and dy each a weight m, and x and dx each a weight of 1.

The "weights" referred to are the same as those for first order isobaric equations so I follow that part. What I don't understand is the process used to determine the weights of each term in an isobaric ODE. For example:

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0$$..eq1.

The book says that from left to right, the "weights" of each term on the LHS are m+1, m+1, 2m, 2m, m + 1.

From an earlier section, an example of a first order isobaric equation is:

$$\frac{{dy}}{{dx}} = - \frac{1}{{2yx}}\left( {y^2 + \frac{2}{x}} \right)$$...eq2

Weights are assigned to y, dy, x and dx as before. However, unlike with eq1, to determine the weights of y, dy, x and dx, eq2 is rewritten as $$\left( {y^2 + \frac{2}{x}} \right)dx + 2yxdy = 0$$ and then the weights are determined.

I would've thought that the method used(as in move everything to LHS) to determine the weights of the terms in eq1, would be applicable to eq2. After all, isn't eq2 just a 'special case' of equations like eq1? I'm just wondering why eq1 can be left as it is in determining the weights of y, dy, dx and x whereas eq2 needs to be rearranged prior to the assignment of weights. Any help would be good, thanks.

Edit: Fixed first paragraph. The terms x and dx are assigned weights of 1 not m.

 

[Nate810]

The reason why you need to rearrange equation 2 is because it is not written in the standard form of an isobaric equation. In an isobaric equation, all the terms that contain derivatives of y should be on the same side and all the terms without derivatives of y should be on the other side. Therefore, in order to assign weights, equation 2 needs to be rearranged so that it is in the standard form. Once it is in the standard form, you can assign weights to each term based on the number of derivatives of y in the term.

 

[thepatientmental]

The process used to determine the weights of terms in an isobaric ODE is based on the principle of dimensional consistency. This means that for an equation to be valid, the units of each term on both sides of the equation must be the same. In order to achieve this, we assign weights to each variable and derivative in the equation, such that when multiplied together, they result in the desired units.

In the first example (eq1), the weights are determined by looking at each term individually and assigning weights based on the units of that term. For example, the first term on the left hand side, x^3(dy/dx)^2, has units of (length)^3 * (1/length)^2 = length. To make it dimensionally consistent, we assign weights of m+1 to y and dy, and a weight of 1 to x and dx. This results in a total weight of (m+1) * (1/length)^3 * (1/length)^2 = 1, as desired.

In the second example (eq2), the equation is already in the form of a first order isobaric equation, where the terms are grouped together with their corresponding derivatives. In this case, we can simply look at each group and assign weights accordingly. For example, the first group, (y^2 + 2/x)dx, has units of (1/length)^2 * length = 1. Therefore, we assign weights of 1 to y and dx, and a weight of 2 to x. This results in a total weight of (1/length)^2 * 1 * length = 1, as desired.

In summary, the method used to determine weights in isobaric ODEs is based on the principle of dimensional consistency. The specific process may vary depending on the form of the equation, but the end goal is to assign weights to each term such that the units on both sides of the equation are the same.